Germany - Math Olympiad Problem | Be Careful!

You should know this approach. Many goes WRONG! Solution

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[franciscook]

Euler's formula: e^iθ = cosθ + i sinθ
a^4=(a-1)^4
(a/(a-1))^4 = 1 = e^(2niπ) for n = 0, 1, 2, 3, ...
take fourth root
a/(a-1)= e^(iπn/2) = 1, i, -1, -i
giving no solution for 1 (it is an eqn of order 3, after all) but the others are
a= i/(1-i) = (1+i)/2
a= 1/(1+1) = 1/2
a= -i/(1+i) = (1-i)/2

[davidsousaRJ]

Just from looking the equation, imediately I knew that a = 1/2 should be one solution. That's because the only way a number to the fourth power to be equal to this number minus one to the fourth power is if the number minus one is the same number with switched sign, that is, x^4 = (-x)^4, therefore a-1 = - a, and a = 1/2. Also, I noticed that the fourth power terms would cancel out on both sides of the equation, and the remaining equation would be a cubic. The other two solutions should be irrational, because there is no other possible real solution to a^4 = (a-1)^4, as I said above. So the only calculation needed is to divide the polynomial "a^4 - (a-1)^4 " by (a-1/2) and solve the resulting quadratic equation. Dividing - 4a³ +6a² - 4a + 1 by a - 1/2 gives a² - a + 1/2, and finding the roots gives a = (1+- i)/2. Very simple problem.

[joelganesh]

You can take fourth roots on both sides, but the catch is that the results are only equal to each other when multiplied with one of the fourth roots of unity. So suppose that a^4 = (a-1)^4, then a = (a-1)*z, for some z such that z^4 = 1. Then, a(1-z) = -z, so a = z/(z-1). The fourth roots of unity are 1, -1, i and -i, so just plugging in these values for z you obtain the answers. Note that z = 1 does not give a solution because you are dividing by 0.

[janda]

My method:

Since a number and its negative raised to an even power give the same solution, a = -(a-1) ==> a = 1/2

Expand the RHS

Cancel a^4

Factor out (a-1/2) using long division

Solve the other factor, a second order polynomial

Get answers a = (1 +/- i) / 2

[gary.h.turner]

Question: Solve for the value of a.
Answer: x = ...

Wait... what?!!! 😱🤔

[KipIngram]

First expand the right hand side and cancel the a^4 term:

a^4 = (a-1)^4
a^4 = a^4 - 4*x^3 + 6*x^2 - 4*x + 1
x^3 - 1.5*x^2 + x - 0.25 = 0

Now recognize that x = 0.5 is a root of the original equation, by inspection. So we can divide out (x - 0.5), which leaves

a^2 - a + 2 = 0

The roots of that are 0.5 +/- j*0.5.

[KrasBadan]

a⁴=(a-1)⁴
Obviously a=½ is a solution.

4a³-6a²+4a-1=0
By dividing it by (2a-1) we get
(2a-1)(2a²-2a+1)=0

D=4-8=-4<0
No other real solutions

(2±2i)/4=½±½i

Solutions are a=½, ½±½i

[leibenzon]

Much simpler solution. Make a change b = a - 1/2. Then you have (b-1/2)^4 = (b+1/2)^4. Open the binomial, and only the odd powers won't cancel out. You end up with 4b^3 + 4b = 0, which solutions are b=0, b=i/2, b=-i/2. Substitute back to a and voila, you have your answer.

[pat]

You can do that very quickly ! a^4 = (a-1)^4 means : a = a - 1 or a = -(a-1) or a = i(a - 1) or a = -i(a-1)
1st equation : no solution and the 3 others are very easy to resolve

[renesperb]

A short way to the three solutions is as follows : you have the possible cases : i*x = x-1, -x= x-1, -i*x = x-1, which leads to the three solutions (1+i)/2, (1 - i)/2, 1/2 .

[mr.soundguy]

a^4=(a-1)^4 so a = w*(a-1) where w is any 4-th root of unity. Then, a*(w-1)=w so a = 1+1/(w-1). Valid solutions: w=+-i, w=-1

[bgkim]

1)a= -(a-1) so that a = 1/2.
2)If we factorize the given formula, (2a-1)(2a^2 -2a +1)=0. We got the solutions (1+i)/2, (1-i)/2

[johnpaullogan]

if x^2=y^2 then x= plus or minus y. a^2= plus or minus (a-1)^2 a^2=plus or minus (a^2-2a+1) a^2=a^2-2a+1 or a^2=-a^2+2a-1. first you subtract a^2-2a from both sides to get 2a=1 which simplifies to a=1/2. the second becomes 0=-2a^2+2a-1. quadratic formula gives (-2(plus/minus)sqrt(4-8))/-4. goes to (-2(plus/minus)2i)/4 which becomes (1+i)/2 or (1-i)/2

[mikeeisler]

Take the quartic root of each side and you have

+- a = +- (a - 1)

==>

(1) a = a - 1

(2) a = 1 - a

(3) -a = a - 1

(4) -a = 1 - a

Equations 2 and 3 resolve to a=1/2

[Tjen]

You can save yourself a few lines by making it squares in the first place:
(a^2)^2=((a-1)^2)^2
a^2=(a-1)^2 v a^2=-(a-1)^2
a^2=a^2-2a+1 v a^2=-a^2+2a-1
-2a+1=0 v 2a^2-2a+1=0

I think this is far easier for possible students to comprehend, then making use of x^2-y^2=(x+y)(x-y) but thats just my opinion as a math teacher ;)

[throx]

Substitute a = b + 1/2, and solve for b. Answer falls out much more easily.

[aeugh]

my direct impulse:
1. a1 = 0.5
2. a^4 cancels out -> polynomial 3rd order
3. polynomial division by (x-0.5)
4. abc (or pq) formula

(I'm German)

[mea]

Theres no way this is an olympiad problem. This is like algebra ii at most.

[eugnsp]

The problem statement must include specification of a field in which solution are to be sought. There is no reason to silently assume it is C.

[chillywilly]

Be Careful not to change the "a" to an "x" in your answer!